3.89.27 \(\int \frac {e^x (11-11 x+5 x^2+8 x^3+(-11+11 x-8 x^2) \log (4))}{10 x^2} \, dx\)

Optimal. Leaf size=24 \[ \frac {e^x \left (-\frac {11}{2}+4 x\right ) (1+x-\log (4))}{5 x} \]

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Rubi [A]  time = 0.10, antiderivative size = 45, normalized size of antiderivative = 1.88, number of steps used = 9, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2199, 2176, 2194, 2177, 2178} \begin {gather*} \frac {4 e^x x}{5}-\frac {4 e^x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x*(11 - 11*x + 5*x^2 + 8*x^3 + (-11 + 11*x - 8*x^2)*Log[4]))/(10*x^2),x]

[Out]

(-4*E^x)/5 + (4*E^x*x)/5 + (E^x*(5 - 8*Log[4]))/10 - (11*E^x*(1 - Log[4]))/(10*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{10} \int \frac {e^x \left (11-11 x+5 x^2+8 x^3+\left (-11+11 x-8 x^2\right ) \log (4)\right )}{x^2} \, dx\\ &=\frac {1}{10} \int \left (8 e^x x+5 e^x \left (1-\frac {8 \log (4)}{5}\right )-\frac {11 e^x (-1+\log (4))}{x^2}+\frac {11 e^x (-1+\log (4))}{x}\right ) \, dx\\ &=\frac {4}{5} \int e^x x \, dx+\frac {1}{10} (5-8 \log (4)) \int e^x \, dx+\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x^2} \, dx-\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x} \, dx\\ &=\frac {4 e^x x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x}-\frac {11}{10} \text {Ei}(x) (1-\log (4))-\frac {4 \int e^x \, dx}{5}+\frac {1}{10} (11 (1-\log (4))) \int \frac {e^x}{x} \, dx\\ &=-\frac {4 e^x}{5}+\frac {4 e^x x}{5}+\frac {1}{10} e^x (5-8 \log (4))-\frac {11 e^x (1-\log (4))}{10 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 22, normalized size = 0.92 \begin {gather*} \frac {e^x (-11+8 x) (1+x-\log (4))}{10 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(11 - 11*x + 5*x^2 + 8*x^3 + (-11 + 11*x - 8*x^2)*Log[4]))/(10*x^2),x]

[Out]

(E^x*(-11 + 8*x)*(1 + x - Log[4]))/(10*x)

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fricas [A]  time = 0.49, size = 26, normalized size = 1.08 \begin {gather*} \frac {{\left (8 \, x^{2} - 2 \, {\left (8 \, x - 11\right )} \log \relax (2) - 3 \, x - 11\right )} e^{x}}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(2*(-8*x^2+11*x-11)*log(2)+8*x^3+5*x^2-11*x+11)*exp(x)/x^2,x, algorithm="fricas")

[Out]

1/10*(8*x^2 - 2*(8*x - 11)*log(2) - 3*x - 11)*e^x/x

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giac [A]  time = 0.12, size = 35, normalized size = 1.46 \begin {gather*} \frac {8 \, x^{2} e^{x} - 16 \, x e^{x} \log \relax (2) - 3 \, x e^{x} + 22 \, e^{x} \log \relax (2) - 11 \, e^{x}}{10 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(2*(-8*x^2+11*x-11)*log(2)+8*x^3+5*x^2-11*x+11)*exp(x)/x^2,x, algorithm="giac")

[Out]

1/10*(8*x^2*e^x - 16*x*e^x*log(2) - 3*x*e^x + 22*e^x*log(2) - 11*e^x)/x

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maple [A]  time = 0.15, size = 27, normalized size = 1.12

method	result	size
gosper	\(-\frac {{\mathrm e}^{x} \left (16 x \ln \relax (2)-8 x^{2}-22 \ln \relax (2)+3 x +11\right )}{10 x}\)	\(27\)
risch	\(-\frac {{\mathrm e}^{x} \left (16 x \ln \relax (2)-8 x^{2}-22 \ln \relax (2)+3 x +11\right )}{10 x}\)	\(27\)
norman	\(\frac {\left (\frac {11 \ln \relax (2)}{5}-\frac {11}{10}\right ) {\mathrm e}^{x}+\left (-\frac {8 \ln \relax (2)}{5}-\frac {3}{10}\right ) x \,{\mathrm e}^{x}+\frac {4 \,{\mathrm e}^{x} x^{2}}{5}}{x}\)	\(32\)
default	\(-\frac {11 \,{\mathrm e}^{x}}{10 x}+\frac {4 \,{\mathrm e}^{x} x}{5}-\frac {3 \,{\mathrm e}^{x}}{10}-\frac {8 \,{\mathrm e}^{x} \ln \relax (2)}{5}+\frac {11 \ln \relax (2) {\mathrm e}^{x}}{5 x}\)	\(33\)
meijerg	\(-\frac {11}{10 x}-\frac {3}{10}+\frac {11 \ln \relax (x )}{10}+\frac {11 i \pi }{10}+\frac {\frac {11 x}{10}+\frac {11}{10}}{x}-\frac {11 \,{\mathrm e}^{x}}{10 x}-\frac {11 \ln \left (-x \right )}{10}-\frac {11 \expIntegralEi \left (1, -x \right )}{10}-\left (-\frac {8 \ln \relax (2)}{5}+\frac {1}{2}\right ) \left (1-{\mathrm e}^{x}\right )-\left (-\frac {11 \ln \relax (2)}{5}+\frac {11}{10}\right ) \left (\ln \relax (x )+i \pi -\ln \left (-x \right )-\expIntegralEi \left (1, -x \right )\right )-\frac {2 \left (-2 x +2\right ) {\mathrm e}^{x}}{5}+\frac {11 \ln \relax (2) \left (\frac {1}{x}+1-\ln \relax (x )-i \pi -\frac {2 x +2}{2 x}+\frac {{\mathrm e}^{x}}{x}+\ln \left (-x \right )+\expIntegralEi \left (1, -x \right )\right )}{5}\)	\(139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*(2*(-8*x^2+11*x-11)*ln(2)+8*x^3+5*x^2-11*x+11)*exp(x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/10*exp(x)*(16*x*ln(2)-8*x^2-22*ln(2)+3*x+11)/x

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maxima [C]  time = 0.41, size = 44, normalized size = 1.83 \begin {gather*} \frac {4}{5} \, {\left (x - 1\right )} e^{x} + \frac {11}{5} \, {\rm Ei}\relax (x) \log \relax (2) - \frac {8}{5} \, e^{x} \log \relax (2) - \frac {11}{5} \, \Gamma \left (-1, -x\right ) \log \relax (2) - \frac {11}{10} \, {\rm Ei}\relax (x) + \frac {1}{2} \, e^{x} + \frac {11}{10} \, \Gamma \left (-1, -x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(2*(-8*x^2+11*x-11)*log(2)+8*x^3+5*x^2-11*x+11)*exp(x)/x^2,x, algorithm="maxima")

[Out]

4/5*(x - 1)*e^x + 11/5*Ei(x)*log(2) - 8/5*e^x*log(2) - 11/5*gamma(-1, -x)*log(2) - 11/10*Ei(x) + 1/2*e^x + 11/
10*gamma(-1, -x)

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mupad [B]  time = 0.07, size = 19, normalized size = 0.79 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (8\,x-11\right )\,\left (x-\ln \relax (4)+1\right )}{10\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(5*x^2 - 2*log(2)*(8*x^2 - 11*x + 11) - 11*x + 8*x^3 + 11))/(10*x^2),x)

[Out]

(exp(x)*(8*x - 11)*(x - log(4) + 1))/(10*x)

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sympy [A]  time = 0.14, size = 27, normalized size = 1.12 \begin {gather*} \frac {\left (8 x^{2} - 16 x \log {\relax (2 )} - 3 x - 11 + 22 \log {\relax (2 )}\right ) e^{x}}{10 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*(2*(-8*x**2+11*x-11)*ln(2)+8*x**3+5*x**2-11*x+11)*exp(x)/x**2,x)

[Out]

(8*x**2 - 16*x*log(2) - 3*x - 11 + 22*log(2))*exp(x)/(10*x)

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